Lời giải
Ta có:
\(\left\{{}\begin{matrix}x\left(x+y\right)=\dfrac{1}{48}\\y\left(x+y\right)=\dfrac{1}{24}\end{matrix}\right.\) \(\Leftrightarrow x\left(x+y\right)+y\left(x+y\right)=\dfrac{1}{48}+\dfrac{1}{24}\)
\(\Leftrightarrow\left(x+y\right)^2=\dfrac{1}{48}+\dfrac{2}{48}=\dfrac{3}{48}=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=\dfrac{1}{4}\\x+y=-\dfrac{1}{4}\end{matrix}\right.\)
Với \(x+y=\dfrac{1}{4}\) thì:
\(\left\{{}\begin{matrix}x=\dfrac{1}{48}:\dfrac{1}{4}=\dfrac{1}{12}\\y=\dfrac{1}{24}:\dfrac{1}{4}=\dfrac{1}{6}\end{matrix}\right.\)
Với \(x+y=-\dfrac{1}{4}\) thì:
\(\left\{{}\begin{matrix}x=\dfrac{1}{48}:-\dfrac{1}{4}=-\dfrac{1}{12}\\y=\dfrac{1}{24}:-\dfrac{1}{4}=-\dfrac{1}{6}\end{matrix}\right.\)
